Learning Objectives for Chapter 4

4

Continuous Random Variables and Probability Distributions

CHAPTER OUTLINE 4-1 Continuous Random Variables 4-7 Normal Approximation to the 4-2 Probability Distributions and Binomial and Poisson Distributions Probability Density Functions 4-8 Exponential Distribution 4-3 Cumulative Distribution Functions 4-9 Erlang and Gamma Distributions 4-4 Mean and Variance of a 4-10 Weibull Distribution Continuous Random Variable 4-11 Lognormal Distribution 4-5 Continuous Uniform Distribution 4-12 Beta Distribution 4-6 Normal Distribution Chapter 4 Title and Outline 1

Learning Objectives for Chapter 4 After careful study of this chapter, you should be able to do the following: 1. 2. 3. 4. 5. 6. 7. 8. 9.

Determine probabilities from probability density functions. Determine probabilities from cumulative distribution functions, and cumulative distribution functions from probability density functions, and the reverse. Calculate means and variances for continuous random variables. Understand the assumptions for some common continuous probability distributions. Select an appropriate continuous probability distribution to calculate probabilities for specific applications. Calculate probabilities, determine means and variances for some common continuous probability distributions. Standardize normal random variables. Use the table for the cumulative distribution function of a standard normal distribution to calculate probabilities. Approximate probabilities for some binomial and Poisson distributions.

Chapter 4 Learning Objectives © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Continuous Random Variables The dimensional length of a manufactured part is subject to small variations in measurement due to vibrations, temperature fluctuations, operator differences, calibration, cutting tool wear, bearing wear, and raw material changes. This length X would be a continuous random variable that would occur in an interval (finite or infinite) of real numbers. The number of possible values of X, in that interval, is uncountably infinite and limited only by the precision of the measurement instrument. Sec 4-1 Continuos Radom Variables © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Continuous Density Functions Density functions, in contrast to mass functions, distribute probability continuously along an interval. The loading on the beam between points a & b is the integral of the function between points a & b.

Figure 4-1 Density function as a loading on a long, thin beam. Most of the load occurs at the larger values of x. Sec 4-2 Probability Distributions & Probability Density Functions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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A probability density function f(x) describes the probability distribution of a continuous random variable. It is analogous to the beam loading.

Figure 4-2 Probability is determined from the area under f(x) from a to b. Sec 4-2 Probability Distributions & Probability Density Functions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Probability Density Function For a continuous random variable X , a probability density function is a function such that (1)

f  x  0

means that the function is always non-negative.



(2)



f ( x)dx  1

 b

(3) (4)

P  a  X  b    f  x  dx  area under f  x  dx from a to b f  x  0

a

means there is no area exactly at x.

Sec 4-2 Probability Distributions & Probability Density Functions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Histograms A histogram is graphical display of data showing a series of adjacent rectangles. Each rectangle has a base which represents an interval of data values. The height of the rectangle creates an area which represents the relative frequency associated with the values included in the base. A continuous probability distribution f(x) is a model approximating a histogram. A bar has the same area of the integral of those limits.

Figure 4-3 Histogram approximates a probability density function. Sec 4-2 Probability Distributions & Probability Density Functions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Area of a Point If X is a continuous random variable, for any x1 and x2 ,

P  x1  X  x2   P  x1  X  x2   P  x1  X  x2   P  x1  X  x2 

(4-2)

which implies that P  X  x   0. From another perspective: As x1 approaches x2 , the area or probability becomes smaller and smaller. As x1 becomes x2 , the area or probability becomes zero.

Sec 4-2 Probability Distributions & Probability Density Functions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-1: Electric Current Let the continuous random variable X denote the current measured in a thin copper wire in milliamperes (mA). Assume that the range of X is 0 ≤ x ≤ 20 and f(x) = 0.05. What is the probability that a current is less than 10mA? Answer: 10

P  X  10    0.5dx  0.5 0

Another example, 20

P  5  X  20    0.5dx  0.75

Figure 4-4 P(X < 10) illustrated.

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Sec 4-2 Probability Distributions & Probability Density Functions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-2: Hole Diameter Let the continuous random variable X denote the diameter of a hole drilled in a sheet metal component. The target diameter is 12.5 mm. Random disturbances to the process result in larger diameters. Historical data shows that the distribution of X can be modeled by f(x)= 20e-20(x-12.5), x ≥ 12.5 mm. If a part with a diameter larger than 12.60 mm is scrapped, what proportion of parts is scrapped? Answer:

Figure 4-5 P  X  12.60  Sec 4-2 Probability Distributions & Probability Density Functions





20e

20 x 12.5

dx  0.135

12.6

© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Cumulative Distribution Functions The cumulative distribution function of a continuous random variable X is, F  x  P  X  x 

x

 f u  du

for    x  

(4-3)



Sec 4-3 Cumulative Distribution Functions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-3: Electric Current For the copper wire current measurement in Exercise 4-1, the cumulative distribution function (CDF) consists of three expressions to cover the entire real number line. 0 x 13 for a normal random variable with μ = 10 and σ2 = 4. Sec 4-6 Normal distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Empirical Rule P(μ – σ < X < μ + σ) = 0.6827 P(μ – 2σ < X < μ + 2σ) = 0.9545 P(μ – 3σ < X < μ + 3σ) = 0.9973

Figure 4-12 Probabilities associated with a normal distribution – well worth remembering to quickly estimate probabilities. Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Standard Normal Distribution A normal random variable with μ = 0 and σ2 = 1 Is called a standard normal random variable and is denoted as Z. The cumulative distribution function of a standard normal random variable is denoted as: Φ(z) = P(Z ≤ z) = F(z) Values are found in Appendix Table III and by using Excel and Minitab. Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-11: Standard Normal Distribution Assume Z is a standard normal random variable. Find P(Z ≤ 1.50). Answer: 0.93319

Figure 4-13 Standard normal PDF

Find P(Z ≤ 1.53). Find P(Z ≤ 0.02).

Answer: 0.93699 Answer: 0.50398

Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-12: Standard Normal Exercises 1.

P(Z > 1.26) = 0.1038

2.

P(Z < -0.86) = 0.195

3.

P(Z > -1.37) = 0.915

4.

P(-1.25 < 0.37) = 0.5387

5.

P(Z ≤ -4.6) ≈ 0

6.

Find z for P(Z ≤ z) = 0.05, z = -1.65

7.

Find z for (-z < Z < z) = 0.99, z = 2.58

Figure 4-14 Graphical displays for standard normal distributions.

Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Standardizing Suppose X is a normal random variable with mean  and variance  2 .  X  x  Then, P  X  x   P    P Z  z     where Z is a standard normal random variable, and z

x   

(4-11)

is the z-value obtainedby standardizing X.

The probability is obtained by using Appendix Table III with z 

x   

.

Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-14: Normally Distributed Current-1 From a previous example

 9  10 x  10 11  10  P  9  X  11  P     2 2 2    P  0.5  z  0.5 

with μ = 10 and σ = 2 mA, what is the probability that the current  P  z  0.5   P  z  0.5  measurement is between  0.69146  0.30854  0.38292 9 and 11 mA? Using Excel Answer: 0.38292 = NORMDIST(11,10,2,TRUE) - NORMDIST(9,10,2,TRUE)

Figure 4-15 Standardizing a normal random variable. Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-14: Normally Distributed Current-2 Determine the value for which the probability that a current measurement is below this value is 0.98. Answer:

 X  10 x  10  P  X  x  P    2 2   x  10    PZ    0.98 2   z  2.05 is the closest value. z  2  2.05   10  14.1 mA.

Using Excel 14.107 = NORMINV(0.98,10,2)

Figure 4-16 Determining the value of x to meet a specified probability. Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-15: Signal Detection-1 Assume that in the detection of a digital signal, the background noise follows a normal distribution with μ = 0 volt and σ = 0.45 volt. The system assumes a signal 1 has been transmitted when the voltage exceeds 0.9. What is the probability of detecting a digital 1 when none was sent? Let the random variable N denote the voltage of noise.

 N  0 0.9  P  N  0.9   P     P  Z  2  0.45 0.45   1  0.97725  0.02275 Using Excel 0.02275 = 1 - NORMDIST(0.9,0,0.45,TRUE) This probability can be described as the probability of a false detection. Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-15: Signal Detection-2 Determine the symmetric bounds about 0 that include 99% of all noise readings. We need to find x such that P(-x < N < x) = 0.99. N x   x P x  N  x  P     0.45 0.45 0.45   x   x  P Z   P  2.58  Z  2.58  0.45   0.45 x  2.58  0.45   0  1.16

Figure 4-17 Determining the value of x to meet a specified probability. Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-15: Signal Detection-3 Suppose that when a digital 1 signal is transmitted, the mean of the noise distribution shifts to 1.8 volts. What is the probability that a digital 1 is not detected? Let S denote the voltage when a digital 1 is transmitted.

 S  1.8 0.9  1.8  P  S  0.9   P    0.45   0.45  P  Z  2   0.02275 Using Excel 0.02275 = NORMDIST(0.9, 1.8, 0.45, TRUE)

This probability can be interpreted as the probability of a missed signal. Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-16: Shaft Diameter-1 The diameter of the shaft is normally distributed with μ = 0.2508 inch and σ = 0.0005 inch. The specifications on the shaft are 0.2500 ± 0.0015 inch. What proportion of shafts conform to the specifications? Let X denote the shaft diameter in inches. Answer: P 0.2485  X  0.2515   0.2515  0.2508   0.2485  0.2508  P Z  0.0005 0.0005    P  4.6  Z  1.4   P  Z  1.4   P  Z  4.6   0.91924  0.0000  0.91924 Using Excel 0.91924 = NORMDIST(0.2515, 0.2508, 0.0005, TRUE) - NORMDIST(0.2485, 0.2508, 0.0005, TRUE)

Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-16: Shaft Diameter-2 Most of the nonconforming shafts are too large, because the process mean is near the upper specification limit. If the process is centered so that the process mean is equal to the target value, what proportion of the shafts will now conform? Answer: P  0.2485  X  0.2515 0.2515  0.2500   0.2485  0.2500  P Z  0.0005 0.0005    P  3  Z  3  P  Z  3  P  Z  3  0.99865  0.00135  0.99730 Using Excel 0.99730 = NORMDIST(0.2515, 0.25, 0.0005, TRUE) - NORMDIST(0.2485, 0.25, 0.0005, TRUE)

By centering the process, the yield increased from 91.924% to 99.730%, an increase of 7.806% Sec 4-6 Normal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Normal Approximations • The binomial and Poisson distributions become more bell-shaped and symmetric as their means increase. • For manual calculations, the normal approximation is practical – exact probabilities of the binomial and Poisson, with large means, require technology (Minitab, Excel). • The normal is a good approximation for the: – Binomial if np > 5 and n(1-p) > 5. – Poisson if λ > 5. Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Normal Approximation to the Binomial Suppose we have a binomial distribution with n = 10 and p = 0.5. Its mean and standard deviation are 5.0 and 1.58 respectively. Draw the normal distribution over the binomial distribution. The areas of the normal approximate the areas of the bars of the binomial with a continuity correction.

Figure 4-19 Overlaying the normal distribution upon a binomial with matched parameters.

Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

40

Example 4-17: In a digital comm channel, assume that the number of bits received in error can be modeled by a binomial random variable. The probability that a bit is received in error is 10-5. If 16 million bits are transmitted, what is the probability that 150 or fewer errors occur? Let X denote the number of errors. Answer: 150

P  X  150    C x 0

16000000 x

10  1 10  5 x

5 16000000  x

Using Excel 0.2280 = BINOMDIST(150,16000000,0.00001,TRUE)

Can only be evaluated with technology. Manually, we must use the normal approximation to the binomial. Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Normal Approximation Method If X is a binomial random variable with parameters n and p, X  np Z np 1  p 

(4-12)

is approximately a standard normal random variable. To approximate a binomial probability with a normal distribution, a continuity correction is applied as follows:  x  0.5  np   P  X  x   P  X  x  0.5   P  Z   np 1  p    and  x  0.5  np   P  X  x   P  X  x  0.5   P  Z   np 1  p    The approximation is good for np  5 and n 1  p   5. Refer to Figure 4-19 to see the rationale for adding and subtracting the 0.5 continuity correction. Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-18: Applying the Approximation The digital comm problem in the previous example is solved using the normal approximation to the binomial as follows: P  X  150   P  X  150.5    X  160 150.5  160   P   160 1  10 5 160 1  10 5       9.5    PZ    P  0.75104   0.2263 12.6491   Using Excel 0.2263 = NORMDIST(150.5, 160, SQRT(160*(1-0.00001)), TRUE) -0.7% = (0.2263-0.228)/0.228 = percent error in the approximation Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-19: Normal Approximation-1 Again consider the transmission of bits. To judge how well the normal approximation works, assume n = 50 bits are transmitted and the probability of an error is p = 0.1. The exact and approximated probabilities are: P  X  2   C050 0.950  C150 0.1 0.9 49   C250 0.12  0.9 48   0.112   X 5 2.5  5  P  X  2  P    50  0.1 0.9   50 0.1 0.9       P  Z  1.18   0.119 Using Excel 0.1117 = BINOMDIST(2,50,0.1,TRUE) 0.1193 = NORMDIST(2.5, 5, SQRT(5*0.9), TRUE) 6.8% = (0.1193 - 0.1117) / 0.1117 = percent error Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-19: Normal Approximation-2 P  X  8   P  X  9   P  X  8.5  8.5  5    PZ    P  Z  1.65   0.05 2.12   P  X  5   P  4.5  X  5.5  5.5  5   4.5  5  P Z  2.12 2.12    P  0.24  Z  0.24   P  Z  0.24   P  Z  0.24   0.19 Using Excel 0.1849 = BINOMDIST(5,50,0.1,FALSE) 0.1863 = NORMDIST(5.5, 5, SQRT(5*0.9), TRUE) - NORMDIST(4.5, 5, SQRT(5*0.9), TRUE) 0.8% = (0.1863 - 0.1849) / 0.1849 = percent error Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Reason for the Approximation Limits The np > 5 and n(1-p) > 5 approximation rule is needed to keep the tails of the normal distribution from getting outof-bounds. As the binomial mean approaches the endpoints of the range of x, the standard deviation must be small enough to prevent overrun. Figure 4-20 shows the asymmetric shape of the binomial when the approximation rule is not met.

Figure 4-20 Binomial distribution is not symmetric as p gets near 0 or 1.

Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Normal Approximation to Hypergeometric Recall that the hypergeometric distribution is similar to the binomial such that p = K / N and when sample sizes are small relative to population size. Thus the normal can be used to approximate the hypergeometric distribution also. hypergeometric distribution

≈

binomial distribution

≈

normal distribution

n / N < 0.1

np < 5 n (1-p ) < 5 Figure 4-21 Conditions for approximatine hypergeometric and binomial with normal probabilities Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Normal Approximation to the Poisson If X is a Poisson random variable with E  X    and V  X    , Z

X 



(4-13)

is approximately a standard normal random variable. The same continuity correction used for the binomial distribution can also be applied. The approximation is good for  5

Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-20: Normal Approximation to Poisson Assume that the number of asbestos particles in a square meter of dust on a surface follows a Poisson distribution with a mean of 100. If a square meter of dust is analyzed, what is the probability that 950 or fewer particles are found? e10001000 x P  X  950    x! x 0 950

... too hard manually!

950.5  1000    P  X  950.5  P  Z   1000    P  Z  1.57   0.058 Using Excel 0.0578 = POISSON(950,1000,TRUE) 0.0588 = NORMDIST(950.5, 1000, SQRT(1000), TRUE) 1.6% = (0.0588 - 0.0578) / 0.0578 = percent error Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Exponential Distribution • The Poisson distribution defined a random variable as the number of flaws along a length of wire (flaws per mm). • The exponential distribution defines a random variable as the interval between flaws (mm’s between flaws – the inverse). Let X denote the number of flaws in x mm of wire. If the mean number of flaws is  per mm, N has a Poisson distribution with mean  x. P  X  x  =P  N  0  

e

 x

 x

0

 e x

0! F  x   P  X  x   1  e   x , x  0, the CDF. Now differentiating: f  x    e   x , x  0, the PDF. Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Exponential Distribution Definition The random variable X that equals the distance between successive events of a Poisson process with mean number of events λ > 0 per unit interval is an exponential random variable with parameter λ. The probability density function of X is:

f(x) = λe-λx for 0 ≤ x <

Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

(4-14)

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Exponential Distribution Graphs The y-intercept of the exponential probability density function is λ. The random variable is nonnegative and extends to infinity. F(x) = 1 – e-λx is well-worth committing to memory – it is used often. Figure 4-22 PDF of exponential random variables of selected values of λ. Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Exponential Mean & Variance If the random variable X has an exponential distribution with parameter  , 1 1 2   EX   and   V  X   2





(4-15)

Note that, for the: • Poisson distribution, the mean and variance are the same. • Exponential distribution, the mean and standard deviation are the same. Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-21: Computer Usage-1 In a large corporate computer network, user log-ons to the system can be modeled as a Poisson process with a mean of 25 log-ons per hour. What is the probability that there are no logons in the next 6 minutes (0.1 hour)? Let X denote the time in hours from the start of the interval until the first log-on. P  X  0.1 



25 x 25 e dx  e 

25 0.1

0.1

 1  F  0.1  0.082 Using Excel 0.0821 = 1 - EXPONDIST(0.1,25,TRUE) Figure 4-23 Desired probability. Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

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Example 4-21: Computer Usage-2 Continuing, what is the probability that the time until the next log-on is between 2 and 3 minutes (0.033 & 0.05 hours)? P  0.033  X  0.05  

0.05



25e 25 x

0.033

 e

 25 x 0.05 0.033

 0.152

 F  0.05   F  0.033  0.152 Using Excel 0.148 = EXPONDIST(3/60, 25, TRUE) - EXPONDIST(2/60, 25, TRUE) (difference due to round-off error) Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

55

Example 4-21: Computer Usage-3 • Continuing, what is the interval of time such that the probability that no log-on occurs during the interval is 0.90?

P  X  x   e25 x  0.90,  25x  ln  0.90  0.10536 x  0.00421 hour  0.253 minute 25 • What is the mean and standard deviation of the time until the next log-in? 1 1    0.04 hour  2.4 minutes  25 1 1    0.04 hour  2.4 minutes  25 Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

56

Characteristic of a Poisson Process • The starting point for observing the system does not matter. • The probability of no log-in in the next 6 minutes [P(X > 0.1 hour) = 0.082], regardless of whether: – A log-in has just occurred or – A log-in has not occurred for the last hour.

• A system may have different means: – High usage period , e.g., λ = 250 per hour – Low usage period, e.g., λ = 25 per hour Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

57

Example 4-22: Lack of Memory Property • Let X denote the time between detections of a particle with a Geiger counter. Assume X has an exponential distribution with E(X) = 1.4 minutes. What is the probability that a particle is detected in the next 30 seconds? P  X  0.5  F  0.5  1 e0.5 1.4  0.30

Using Excel 0.300 = EXPONDIST(0.5, 1/1.4, TRUE)

• No particle has been detected in the last 3 minutes. Will the probability increase since it is “due”? P  X  3.5 X  3 

P  3  X  3.5 F  3.5  F  3 0.035    0.30 P  X  3 1  F  3 0.117

– No, the probability that a particle will be detected depends only on the interval of time, not its detection history. Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

58

Lack of Memory Property • • • • • • •

Areas A+B+C+D=1 A = P(X < t2) A+B+C = P(Xt1) C/(C+D) = P(Xt1) A = C/(C+D)

Figure 4-24 Lack of memory property of an exponential distribution.

Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

59

Exponential Application in Reliability • The reliability of electronic components is often modeled by the exponential distribution. A chip might have mean time to failure of 40,000 operating hours. • The memoryless property implies that the component does not wear out – the probability of failure in the next hour is constant, regardless of the component age. • The reliability of mechanical components do have a memory – the probability of failure in the next hour increases as the component ages. The Weibull distribution is used to model this situation.

Sec 4-8 Exponential Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

60

Erlang & Gamma Distributions • The Erlang distribution is a generalization of the exponential distribution. • The exponential models the interval to the 1st event, while the Erlang models the interval to the rth event, i.e., a sum of exponentials. • If r is not required to be an integer, then the distribution is called gamma. • The exponential, as well as its Erlang and gamma generalizations, is based on the Poisson process. Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

61

Example 4-23: Processor Failure The failures of CPUs of large computer systems are often modeled as a Poisson process. Assume that units that fail are repaired immediately and the mean number of failures per hour is 0.0001. Let X denote the time until 4 failures occur. What is the probability that X exceed 40,000 hours? Let the random variable N denote the number of failures in 40,000 hours. The time until 4 failures occur exceeds 40,000 hours iff the number of failures in 40,000 hours is ≤ 3. P  X  40, 000   P  N  3 E  N   40, 000  0.0001  4 failure in 40,000 hours e4 4k P  N  3    0.433 k! k 0 3

Using Excel 0.433 = POISSON(3, 4, TRUE)

Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

62

Erlang Distribution Generalizing from the prior exercise: r 1

P  X  x  

e

x

k 0

 x k!

k

 1 F  x

Now differentiating F  x  : f  x 

 r x r 1e  x

 r  1!

for x  0 and r  1, 2,...

Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

63

Gamma Function The gamma function is the generalization of the factorial function for r > 0, not just non-negative integers. 

  r    x r 1e  x dx, for r  0

(4-17)

0

Properties of the gamma function   r    r  1   r  1 recursive property   r    r  1 !

factorial function

 1  0!  1  1 2    1 2  1.77

useful if manual

Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

64

Gamma Distribution The random variable X with a probability density function: f  x 

 r xr 1e x r 

, for x  0

(4-18)

has a gamma random distribution with parameters λ > 0 and r > 0. If r is an positive integer, then X has an Erlang distribution. Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

65

Mean & Variance of the Gamma • If X is a gamma random variable with parameters λ and r, μ = E(X) = r / λ and σ2 = V(X) = r / λ2

(4-19)

• r and λ work together to describe the shape of the gamma distribution.

Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

66

Gamma Distribution Graphs The λ and r parameters are often called the “shape” and “scale”, but may take on different meanings. Different parameter combinations change the distribution. The distribution becomes symmetric as r (and μ) increases.

Name Text Excel Minitab Scale λ β=1/λ 1/λ Shape r α r

Figure 4-25 Gamma probability density functions for selected values of λ and r.

Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

67

Example 4-24: Gamma Application-1 The time to prepare a micro-array slide for high-output genomics is a Poisson process with a mean of 2 hours per slide. What is the probability that 10 slides require more than 25 hours? Let X denote the time to prepare 10 slides. Because of the assumption of a Poisson process, X has a gamma distribution with λ = ½, r = 10, and the requested probability is P(X > 25). Using the Poisson distribution, let the random variable N denote the number of slides made in 10 hours. The time until 10 slides are made exceeds 25 hours iff the number of slides made in 25 hours is ≤ 9. P  X  25  P  N  9  E  N   25 1 2   12.5 slides in 25 hours e12.5 12.5 P  N  9    0.2014 k ! k 0 k

9

Using Excel 0.2014 = POISSON(9, 12.5, TRUE)

Using the gamma distribution, the same result is obtained. 0.510 x9e0.5 x P  X  25  1   dx  10 0 25

Using Excel 0.2014 = 1 - GAMMADIST(25,10,2,TRUE)

Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

68

Example 4-24: Gamma Application-2 What is the mean and standard deviation of the time to prepare 10 slides? r

10 EX     20 hours  0.5 r

10 V X   2   40 hours 2  0.25 SD  X  

10



 40  6.32 hours

Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

69

Example 4-24: Gamma Application-3 The slides will be completed by what length of time with 95% probability? That is: P(X ≤ x) = 0.95 Distribution Plot

Gamma, Shape=10, Scale=2, Thresh=0 0.07

0.95

0.06

Density

0.05 0.04 0.03 0.02 0.01 0.00

0

X

31.4

Minitab: Graph > Probability Distribution Plot > View Probability

Using Excel 31.41 = GAMMAINV(0.95, 10, 2)

Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

70

Chi-Squared Distribution • The chi-squared distribution is a special case of the gamma distribution with – λ = 1/2 – r = ν/2 where ν (nu) = 1, 2, 3, … – ν is called the “degrees of freedom”.

• The chi-squared distribution is used in interval estimation and hypothesis tests as discussed in Chapter 7.

Sec 4-9 Erlang & Gamma Distributions © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

71

Weibull Distribution • The Weibull distribution is often used to model the time until failure for physical systems in which failures: – Increase over time (bearings) – Decrease over time (some semiconductors) – Remain constant over time (subject to external shock)

• Parameters provide flexibility to reflect an item’s failure experience or expectation. Sec 4-10 Weibull Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

72

Weibull PDF The random variable X with probability density function

  1  x   f  x   x e for x  0 

(4-20)

is a Weibull random variable with scale parameter   0 and shape parameter   0. The cumulative density function is: F  x  1 e

 x  



(4-21)

 1   E  X      1       2  1  2     V  X     1       1            2

2

2

Sec 4-10 Weibull Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

(4-21a) 73

Weibull Distribution Graphs Added slide

Figure 4-26 Weibull probability density function for selected values of δ and β.

Sec 4-10 Weibull Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

74

Example 4-25: Bearing Wear • The time to failure (in hours) of a bearing in a mechanical shaft is modeled as a Weibull random variable with β = ½ and δ = 5,000 hours. • What is the mean time until failure? E  X   5000   1  1 2   5000   1.5  5000  0.5   4, 431.1 hours

Using Excel 4,431.1 = 5000 * EXP(GAMMALN(1.5))

• What is the probability that a bearing will last at least 6,000 hours? (error in text solution) P  X  6, 000   1  F  6, 000   e e

1.0954

 0.334

 6000     5000 

0.5

Using Excel 0.334 = 1 - WEIBULL(6000, 1/2, 5000, TRUE)

Sec 4-10 Weibull Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

75

Lognormal Distribution • Let W denote a normal random variable with mean of θ and variance of ω2, i.e., E(W) = θ and V(W) = ω2 • As a change of variable, let X = eW = exp(W) and W = ln(X) • Now X is a lognormal random variable. F  x   P  X  x   P exp W   x   P W  ln  x   ln  x  -θ    ln  x  -θ  =P  Z   =Φ    for x  0 ω    ω   0 for x  0 f  x 

1 x 2

EX   e

  2 2

e

 ln  x       2 

and

2

for 0  x   V X   e

2  2

e

2



1

Sec 4-11 Lognormal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

(4-22) 76

Lognormal Graphs

Figure 4-27 Lognormal probability density functions with θ = 0 for selected values of ω2. Sec 4-11 Lognormal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

77

Example 4-27: Semiconductor Laser-1 The lifetime of a semiconductor laser has a lognormal distribution with θ = 10 and ω = 1.5 hours. • What is the probability that the lifetime exceeds 10,000 hours?

P  X  10, 000   1  P exp W   10, 000   1  P W  ln 10, 000    ln 10, 000   10   1    1.5    1    0.5264   0.701

1 - NORMDIST(LN(10000), 10, 1.5, TRUE) = 0.701 Sec 4-11 Lognormal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

78

Example 4-27: Semiconductor Laser-2 • What lifetime is exceeded by 99% of lasers? P  X  x   P  exp W   x   P W  ln  x    ln  x   10   1     0.99 1.5    1    z   0.99 therefore z  2.33 ln  x   10  2.33 and x  exp  6.505   668.48 hours 1.5

-2.3263 = NORMSINV(0.99) 6.5105 = -2.3263 * 1.5 + 10 = ln(x) 672.15 = EXP(6.5105) (difference due to round-off error)

• What is the mean and variance of the lifetime? E  X   e 

2

2

 e101.5

2

2

 exp 11.125  67,846.29









V  X   e 2  e  1  e 2101.5 e1.5  1 2

2

2

2

 exp  22.25  exp  2.25   1  39, 070, 059,886.6

SD  X   197, 661.5

Sec 4-11 Lognormal Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

79

Beta Distribution A continuous distribution that is flexible, but bounded over the [0, 1] interval is useful for probability models. Examples are: – Proportion of solar radiation absorbed by a material. – Proportion of the max time to complete a task. The random variable X with probability density function       1  1 f  x = x 1  x  for 0  x  1         is a beta random variable with parameters   0 and   0.

Sec 4-12 Beta Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

80

Beta Shapes Are Flexible Distribution shape guidelines: 1. If α = β, symmetrical about x = 0.5. 2. If α = β = 1, uniform. 3. If α = β < 1, symmetric & U- shaped. 4. If α = β > 1, symmetric & mound-shaped. 5. If α ≠ β, skewed.

Figure 4-28 Beta probability density functions for selected values of the parameters α and β.

Sec 4-12 Beta Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

81

Example 4-27: Beta Computation-1 Consider the completion time of a large commercial real estate development. The proportion of the maximum allowed time to complete a task is a beta random variable with α = 2.5 and β = 1. What is the probability that the proportion of the max time exceeds 0.7? Let X denote that proportion. P  X  0.7  

      1  1 x 1  x 0.7           dx 1

  3.5     2.5    1

1



x1.5 dx

0.7

2.5 1.5  0.5   x 2.5  1.5  0.5   1 2.5  1   0.7 

2.5

 0.59

1

0.7

Using Excel 0.590 = 1 - BETADIST(0.7,2.5,1,0,1)

Sec 4-12 Beta Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

82

Example 4-27: Beta Computation-2 This Minitab graph illustrates the prior calculation. FIX Example 4-28

Beta: alpha=2.5, beta=1 2.5 0.590

Density

2.0

1.5

1.0

0.5

0.0

0

X

0.7

1

Sec 4-12 Beta Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

83

Mean & Variance of the Beta Distribution If X has a beta distribution with parameters α and β,   EX    2 V X  

    2         1

Example 4-28: In the prior example, α = 2.5 and β = 1. What are the mean and variance of this distribution?  2 

2.5 2.5   0.71 2.5  1 3.5 2.5 1

 2.5  1  2.5  1  1 2



2.5  0.045 2 3.5  4.5 

Sec 4-12 Beta Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

84

Mode of the Beta Distribution If α >1 and β > 1, then the beta distribution is mound-shaped and has an interior peak, called the mode of the distribution. Otherwise, the mode occurs at an endpoint. Distribution Plot

Beta Distribution: alpha=2.5 beta 1 1.1

2.5

2.0

Density

General formula:  1 Mode    2 for   0 and   0.

1.5

1.0

0.5

0.0

0.0

0.2

0.4

0.6

0.8

1.0

X

case alpha beta mode Example 4-28 2.25 1 1.00 = (2.5-1) / (2.5+1.0-2) Alternate 2.25 1.1 0.94 = (2.5-1) / (2.5+1.1-2) Sec 4-12 Beta Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

85

Extended Range for the Beta Distribution The beta random variable X is defined for the [0, 1] interval. That interval can be changed to [a, b]. Then the random variable W is defined as a linear function of X:

W = a + (b –a)X With mean and variance: E(W) = a + (b –a)E(X) V(W) = (b-a)2V(X)

Sec 4-12 Beta Distribution © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.

86

Important Terms & Concepts of Chapter 4 Beta distribution Mean of a function of a continuous random variable Chi-squared distribution Normal approximation to binomial Continuity correction & Poisson probabilities Continuous uniform distribution Cumulative probability distribution Normal distribution Probability density function for a continuous random variable Probability distribution of a continuous random variable Erlang distribution Standard deviation of a continuous Exponential distribution random variable Gamma distribution Standardizing Lack of memory property of a Standard normal distribution continuous random variable Variance of a continuous random Lognormal distribution variable Mean for a continuous random Weibull distribution variable Chapter 4 Summary

87 © John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.